Calculate all the generators in $\mathbb{Z}/61$

Question

Here's the exact question I trying answer: "Calculate all the generators in $(\mathbb{Z}/61)^\times$ . You may assume that $g = 2$ is one such generator. " Does this question mean calculate the number of elements that are generators, or calculate each individual generator? I can't imagine that it is asking to count the number of them because that would just be $61-1$ which is $60$, and that $g=2$ was mentioned just to confuse things. But could someone please explain how I could calculate each of the generators, using the knowledge that 2 is one them. I've looked around online, but I'm stumped. I assume there's some method for doing this other than just slow going through each element one at a time?

Answer 1

Hint: $\text{ord}(a^k) = \dfrac{\text{ord}(a)}{\gcd(\text{ord}(a), k)}$

Answer 2

Hints for you to work out:

== $\;\Bbb Z/61\Bbb Z=:\Bbb F_{61}\;$ is a field .

== A finite subgroup of the multiplicative group of any field is cyclic .

== In a cyclic group of order $\;n\;$ there are exactly $\;\varphi(n)\;$ generators of that group (i.e., elements of order $\;n$) , with $\;\varphi=$ Euler's Totient Function

== If $\;G\;$ is a cyclic group of order $\;n\;$ and $\;G=\langle\,x\,\rangle\;$ , then all the generators of the group are given by

$$\;\{\,x^k\;:\;\;\text{g.c.d.}\,(k,n)=1\,\}$$